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\begin{document}
\title{格点QCD基本公式}
\author{李衢智}
\date{2022-11-16}
\maketitle
\pagenumbering{roman}
\tableofcontents
\newpage
\pagenumbering{arabic}


\section{Wilson actions}
Pure gauge:
\begin{formalred}
  \begin{equation}
	U_{\mu\nu}(n) = U_\mu(n) U_\nu( n + \hat \nu ) U_\mu(n + \hat
	\mu)^\dagger U_\nu(n)^\dagger.
  \end{equation}
  \begin{equation}
	S_G[U] = \frac{\beta}{N} \sum_{n\in\Lambda } \sum_{\mu < \nu}
	\mathcal{Re}tr[1 - U_{\mu\nu}(n)  ] 
  \end{equation}
\end{formalred}

The fermion lattice action is a sum over $N_f$ flavors:
\begin{formalbrown}
	\begin{equation}
	  S_F[\psi, \bar \psi, U] = \sum^{f = 1}_{N_f} a^4 \sum^{n,m \in
	  \Lambda }_{  } \bar\psi^f(n) D^f(n|m)\psi^f(m).
	\end{equation}
\end{formalbrown}
and the lattice Dirac operator is given by :
\begin{formalgreen}
	\begin{equation}
	  D^f(n|m)_{\alpha,\beta;a,b} = \left( m^f + \frac{4}{a} \right)
	  \delta_{\alpha\beta}\delta_{ab}\delta_{n,m} - \frac{1}{2a}
	  \sum^{\mu = \pm 1 }_{\pm 4}( 1 - \gamma_\mu )_{\alpha\beta}
	  U_\mu(n)_{ab}\delta_{n + \hat \mu, m}. 
	\end{equation}
\end{formalgreen}
We use the conventions:
\begin{equation}
  \gamma_{-\mu} = - \gamma_\mu, \quad U_{-\mu}(n) = U_\mu(n - \hat
  \mu), \quad \mu = 1, 2, 3, 4.
\end{equation}

We remark that Wilson's Dirac operator is $\gamma_5$- hermitian, i.e.,
it obeys :
\begin{equation}
	\gamma_5 D \gamma_5 = D^\dagger.
\end{equation}
\subsection{Wilson-Clover action}
For $O(a)$ improvement of the Wilson lattice action it is sufficient
to add the Pauli term:
\begin{formalbrown}
	\begin{equation}
S_I = S_{Wilson}  + c_{sw} a^5 \sum_{n\in \Lambda}
\sum_{\mu < \nu} \bar \psi(n) \frac{1}{2}
\sigma_{\mu\nu}\hat F_{\mu\nu}(n)\psi(n).
	\end{equation}
\end{formalbrown}
where :
\begin{equation}
  \hat F_{\mu\nu}(n)= \frac{-i}{8a^2}(Q_{\mu\nu}(n) - Q_{\nu\mu}(n)),
\end{equation}
and the $Q_{\mu\nu}(n)$ is the sum of plaquettes $U_{\mu,\nu}(n)$ :
\begin{equation}
  Q_{\mu\nu}(n) = U_{\mu,\nu}(n) + U_{\nu,-\mu}(n) +
  U_{-\mu,-\nu}(n) + U_{i\nu,\mu}(n)
\end{equation}
\subsection{Hybrid Monte Carlo}
The force term is :
\begin{equation}
  F[U,\phi] = \sum_{i = 1}^{8} T_i \nabla^i (S_G[U] +
  \phi\dagger(DD\dagger)^{-1}\phi)\quad \in SU(3). 
\end{equation}
The force from the Wilson gauge action:
\begin{formalbrown}
  \begin{equation}
	\nabla^i tr[UA + A^\dagger U^\dagger] = tr[i T_i U A - i A^\dagger
	U^\dagger T_i] = tr[i T_i(UA - A^\dagger U^\dagger)] = - \frac{\beta}{12} i (UA - A^\dagger U^\dagger).
	\end{equation}
\end{formalbrown}
where $A$ denotes the sum of adjacent staples:
\begin{formalbrown}
\begin{equation}
  A = \sum_{\nu \ne \mu} U_\nu(n + \hat \mu) U_{-\mu}(n + \hat \mu +
  \hat \nu)U_{-\nu}(n + \hat \nu) + U_{- \nu }(n + \hat \mu) U_{-\mu}(n + \hat \mu -
  \hat \nu)U_{\nu}(n - \hat \nu)
\end{equation}	
\end{formalbrown}
The contribution of the fermion action to the force is :
\begin{formalbrown}
	\begin{align}
		  \nabla^i \left( \phi^\dagger(D D^\dagger)^{-1} \phi \right) =& -
\phi^\dagger (D D ^\dagger)^{-1} ( \nabla^i D D ^\dagger )(D D
^\dagger)^{-1} \phi \\
& = - 
\phi^\dagger (D D ^\dagger)^{-1} (\frac{\partial D}{\partial \omega^i}
D ^\dagger + D \frac{\partial D ^\dagger}{\partial \omega^i} 
+ )(D D
^\dagger)^{-1} \phi 
	\end{align}
\end{formalbrown}

\subsection{Wilson flow}
The Wilson flow is defined by the equations:
\begin{equation}
  \dot V_{t}(x,\mu) = - g_0^2\{\partial_{x,\mu}S_G(V_t)\}V_t(x,\mu),
  \quad V_t(x,\mu)|_{t=0} = U_\mu(x).
\end{equation}

\section{Computation of quark propagators}
Computation of quark propagators amounts to solving The Dirac
equation :
\begin{equation}
	D \psi(x) = \eta(x) .
\end{equation}
This equation can only be solved iteratively, and practical measure
for the accuracy of an approximate solution $\phi$ is the norm of the
associated residue:
\begin{equation}
	\rho = \eta -D \phi .
\end{equation}
If, say, $||\rho ||$ for some small value $\epsilon$. The residue is
found to be bounded by:
\begin{formalred}
	\begin{equation}
  || \psi -\phi || < \epsilon \kappa(D)  || \psi ||
	\end{equation}
\end{formalred}
where 
\begin{equation}
  \kappa(D) = ||D|| || D^{-1}||.
\end{equation}
The condition number :
\begin{formalbrown}
	\begin{equation}
	  \kappa (D) = (\alpha_{max}/ \alpha_{min}) \propto(am)^{-1} 
	\end{equation}
\end{formalbrown}
\subsection{Even-odd preconditioned}
Given a matrix $M$ written in block form:
\begin{equation}
	\begin{bmatrix} 
	  A_{E,E} & D_{E,O}\\
	  D_{O,E} & A_{OO}
	\end{bmatrix} 
\end{equation}
The preconditioning consists of using the triangular matrices:
\begin{equation}
  L = \begin{bmatrix} 1 & 0 \\
   D_{O,E} A^{-1}_{E,E}& 1
 \end{bmatrix} 
\end{equation}
and
\begin{equation}
  U = \begin{bmatrix} 1 & A^{-1}_{E,E} D_{E,O} \\
   0& 1
 \end{bmatrix} 
\end{equation}
The preconditioned matrix is formed from:
\begin{equation}
  M^\prime = L^{-1}MU^{-1}.
\end{equation}
\section{Choosing valence and sea quark to be different}
Consider the two-point function for a charged pion:
\begin{equation}
  \langle\pi ^+(x) \pi^-(y)\rangle = - \frac{1}{Z} \int \prod_\mu
  [dU_\mu]Det[\slashed{D} +M]tr[\gamma_5(\slashed{D} +m_u)^{-1}(y,x)
  \gamma_5 (\slashed{D} + m_d)^{-1}(x,y)]
\end{equation}
Nothing stops us from the taking the quark mas  in the propagators
unequal to those in the determinant, or, even more drastically, from
choosing a completely different discretization for $\slashed{D}$ in
the propagators and in the determinant!
\section{Jack-knife Resampling}
Original Datasets:
\begin{itemize}
  \item cnfg 1: $\vec{C}^1= (C^1_0,\dots, C^1_t)$;
  \item $\dots\dots$;
  \item cnfg N: $\vec{C}^N= (C^N_0,\dots, C^N_t)$;
\end{itemize}
dropping chfg $i$, calculating mean value generate $i$th Jack-knife
sample:
\begin{itemize}
  \item $\tilde C^i = \frac{N}{N-1}\bar C - \frac{1}{N-1} \vec{C}^i$ 
\end{itemize}
The mean value of the new samples is equal to the old samples:
\begin{equation}
	\bar C = \frac{1}{N} \sum_i \vec C^i = \frac{1}{N} \sum_i \tilde
	C^i
\end{equation}
The  covariance:



\begin{equation}
  \rho = (N-1)\times \tilde \rho,\quad    \rho_{t,t^\prime} =
  \frac{1}{N-1}\sum_i(C^i_t - \bar C_t)(C^i_{t^\prime} - \bar
  C_{t^\prime})
\end{equation}
And the variance:
\begin{equation}
  \sigma = \sqrt{N-1} \times \tilde \sigma.
\end{equation}
\subsection{Renormalization and lattice artifacts}
For a free scalar field theory on an infinite 4-dimensional hypercubic
lattice with standard action:
\begin{equation}
  S_0 = a^4 \sum_{x,\mu} \frac{1}{2}[\partial_\mu \phi(x) \partial_\mu
  \phi(x) + m^2 \phi(x)^2]
\end{equation}
where $\partial_\mu f(x) = [f(x+ a \hat \mu) - f(x)]/a$. The Fourier
transformation is defined:
\begin{equation}
  \tilde f(p) = \sum_x f(x) e^{-i p\cdot x}
\end{equation}
And the inverse transformation:
\begin{equation}
  f(x) = \frac{1}{V} \sum_p \tilde f(p) e^{i p \cdot x}
\end{equation}
\begin{align}
	\label{eq:fre}
	&\sum_{x,\mu} \partial_\mu\phi(x) \partial_\mu \phi(x) =
	\frac{1}{a^2} [\phi(x+ a \hat \mu)^2 - 2 \phi(x+ a \hat
	\mu)\phi(x) + \phi(x)]	\\
	&=\frac{1}{a^2V^2}\sum_x\sum_{p,k}\tilde \phi(p)\tilde \phi(k)[ e^{i p\cdot(x+ \hat \mu)}
	e^{i k\cdot (x + \hat \mu)} - 2 e^{i p \cdot (x+\hat \mu)} e^{i k
	\cdot x} + e^{ip\cdot x}e^{ik\cdot x} ] \\
	&=\frac{1}{a^2V}\sum_{p,k}\tilde \phi(p)\tilde \phi(k)[e^{i\hat \mu\cdot(p+k)}\delta_{p,k} - 2
	e^{i p \cdot \mu}\delta_{p,k} + \delta_{p,k}] \\
	& = \frac{1}{a^2V}\sum_{p}\tilde \phi(p)\tilde \phi(k)(e^{2ip\cdot
	\mu} - 2 e^{i p\cdot \mu} + 1) \\
	& = \frac{1}{a^2V} \sum_{p}\tilde \phi(p)\tilde
	\phi(k)(1-e^{ip\cdot \hat \mu})^2\\
  &=4\sum_{p}\tilde \phi(p)\tilde \phi(k)e^{ip\cdot \hat \mu}(\sin
  \frac{p\cdot \hat \mu}{2})^2
\end{align}
\begin{align}
	\label{eq:s1}
	\partial_\mu^2 \phi(x) &= \partial_\mu[\phi(x + \hat \mu) - \phi(x)] \\
&=[\phi(x + 2\hat \mu) - \phi(x + \hat \mu)  - \phi(x+\hat \mu) + \phi(x)]
\end{align}
\begin{align}
	\label{eq:s2}
	\partial_\mu^2\phi(x) \partial_\mu^2\phi(x) = &	\phi(x + 2
	\hat\mu)^2 +  4 \phi(x + \hat\mu)^2 + \phi(x)^2 \\
 &-4 \phi(x+\hat\mu)\phi(x + 2\hat\mu ) - 4 \phi(x + \hat\mu)\phi(x) +2
 \phi(x)\phi(x + 2\hat\mu) 
\end{align}
\section{数据处理}
假设我们测了一组数据$\{x_1, x_2, \dots, x_n
\}$，现在我们利用这组数据去估计平均值以及标准偏差。我们可以直接估计平均值$\bar
x$为:  \[
\bar x = \frac{1}{N} \sum_i x_i
.\] 
以及标准偏差$\sigma$
 \[
\sigma^2 = \sum_i(x_i - \bar x)
.\] 
我们也可以从另一个角度去看待这组数据，假设有$N$个随机变量
$X_i$，它们的均值和标准偏差都相等:
 \[
   \langle X_i \rangle = \langle X \rangle, \quad, \sigma_{X_i}^2 =
   \langle (X_i - \langle X_i \rangle)^2 \rangle = \sigma_{X}^2
.\] 

然后引入它们的无偏估计：
\begin{equation}
  \hat X = \frac{1}{N} \sum_i X_i,\quad \hat \sigma_{X} =
  \frac{1}{N-1} \sum_i (X_i - \hat X)^2
\end{equation}
这两个随机变量有各自的作用，如果我们估计上面所测数据的平均值，以及所估计平均值的误差，我们可以用$\hat
X$。首先这个随机变量的平均值等于我们第一种方法给出的:
\begin{equation}
	\langle\hat X\rangle =\frac{1}{N} \sum_i \langle X_i \rangle =
	\langle X \rangle
\end{equation}

而这个随机变量的标准偏差为：
\begin{equation}
  \begin{split}
	\sigma_{\hat X}^2 &= \langle (\hat X - \langle X \rangle)^2 = \left\langle
  \left(\frac{1}{N}\sum_i (X_i - \langle X \rangle )\right)^2
\right\rangle \\
&= \frac{1}{N^2} \left\langle\sum_{i,j}(X_i -\langle X \rangle)(X_j 
  - \langle X \rangle) \right \rangle \\
& = \frac{1}{N} \langle X_i^2\rangle - \langle X \rangle^2
+\frac{1}{N^2} \sum_{i\neq j}\langle X_i X_j\rangle
  \end{split}
\end{equation}
如果数据点之间没有相互关联，那么有：
\begin{equation}
  \sigma_{\hat X}^2 = \frac{1}{N} \sigma_{X_i}^2 = \frac{1}{N}
  \sigma_{X}^2
\end{equation}

\begin{equation}
	\begin{split}
	  \left\langle (\tilde X_{i} - \langle\tilde X\rangle)^2
	  \right\rangle &=\left \langle \left(  \frac{N}{(N - 1)}\langle
	  X\rangle - \frac{1}{N -1}X_i - \langle X\rangle\right)^2 \right \rangle \\
  & = \frac{1}{(N - 1)^2}\langle (X_i - \langle X \rangle)^2
  \rangle
	\end{split}
\end{equation}








\end{document}
